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Calculus1 18 Online
OpenStudy (anonymous):

Can someone tell me which limits i shud use for part b to find the area ?

OpenStudy (anonymous):

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OpenStudy (campbell_st):

well you need to solve 2x - 3 = 0 that is the zero of the curve... and based on the diagram, the upper limit... the area is below the curve so remember to use absolute values

OpenStudy (aum):

Lower limit is: x = 0 Upper limit is: x = 3/2

OpenStudy (anonymous):

@aum , @campbell_st why is the lower limit zero ? why isnt it -3 ?

OpenStudy (aum):

For any point on the y-axis, x = 0. The -3 is the y-intercept and not the x-value. The limits of integration goes from the y-axis (that is, x = 0) to where it crosses the x-axis (that is, x = 3/2).

OpenStudy (anonymous):

Oh that was so stupid of me. So thankful @aum . Aprreciate it :)

OpenStudy (aum):

np. You are welcome.

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